BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
Live EA class + 6 months of EA OnDemand
  • Expert-led weekly online sessions
  • EA Masterclass access between classes
View EA class & enroll
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

130-point score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

probability!!!

Expert replies
by kiennguyen » Mon Oct 26, 2009 6:48 pm
try this one! OA later! please clarify your answer!

Tanya prepared four different letters to be sent to four different addresses. For each letter, she prepared an envelope with its correct address. If the 4 letters are to be put into the envelopes at random, what is the probability that only one letter will be put into the envelope with its correct address?
Join the discussion
Source: — Problem Solving |

by capnx » Mon Oct 26, 2009 7:12 pm
1/6?
Join the discussion

Re: probability!!!

by uttam.albela » Mon Oct 26, 2009 9:54 pm
kiennguyen wrote:try this one! OA later! please clarify your answer!

Tanya prepared four different letters to be sent to four different addresses. For each letter, she prepared an envelope with its correct address. If the 4 letters are to be put into the envelopes at random, what is the probability that only one letter will be put into the envelope with its correct address?
Total number of ways of arranging 4 letters into 4 envelopes = 4! = 24

Now calculate the total number of ways of required arranging that is one correct and 3 incorrect.
One correct can be chosen in 4C1=4 ways
3 incorrect with 3 letters and 3 envelopes = 2 ways
Total number of required ways = 4*2=8

Probability required = 8/24 = 1/3
Join the discussion

good job!

by kiennguyen » Tue Oct 27, 2009 4:59 am
OA is 1/3
Join the discussion

Re: probability!!!

by chipbmk » Tue Oct 27, 2009 4:34 pm
Total number of ways of arranging 4 letters into 4 envelopes = 4! = 24

Now calculate the total number of ways of required arranging that is one correct and 3 incorrect.
One correct can be chosen in 4C1=4 ways
3 incorrect with 3 letters and 3 envelopes = 2 ways
Total number of required ways = 4*2=8

Probability required = 8/24 = 1/3[/quote]

Can you please explain the portion I bolded above.

Why do we need to add those two ways?
Join the discussion

Re: probability!!!

by uttam.albela » Tue Oct 27, 2009 7:52 pm
chipbmk wrote:Total number of ways of arranging 4 letters into 4 envelopes = 4! = 24

Now calculate the total number of ways of required arranging that is one correct and 3 incorrect.
One correct can be chosen in 4C1=4 ways
3 incorrect with 3 letters and 3 envelopes = 2 ways
Total number of required ways = 4*2=8

Probability required = 8/24 = 1/3
Can you please explain the portion I bolded above.

Why do we need to add those two ways?[/quote]

When you are selecting a correct set of letter and envelope(Say A+a), there are 2 different ways in which letters BCD can be arranged in bcd envelopes with none of these 3 letters in correct envelope as shown below.

BCD BCD
cdb dbc

So 2 ways when letter A is in correct envelopes and all other 3 letters in wrong envelope.
Similarly, with B in correct envelope, you have 2 ways and so on.

So 2*4 = 8 ways to achieve stated event.

Hopefully, the explanation is clear now.
Join the discussion